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Answer:
The heat of combustion of liquid butane is: -2636 kJ/mol
Explanation:
This problem is solved by using Hess law of heats summation:
We have the heat of combustion for gaseous butane:
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
and we want the heat of combustion for liquid butane.
What we need to do is add the heat of vaporization for liquid butane to this equation:
C₄H₁₀ (l) ⇒ C₄H₁₀ (g) ΔHºvap = 22.44 kJ/mol
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
<u>______________________________________________</u>
C₄H₁₀ (l) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)
So,
ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol
Blood, teeth and are mammals
Answer:
When the concentration of a reactant increases, there will be more chemical present. Due to more reactant particles moving together, more collisions are allowed to happen and with that, the rate of the reaction is increased. So, the higher the concentration of reactants, the faster the reaction rate will be.
Hope this helped you! :)
